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Electrochemistry cathode question |
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| Nov19-12, 06:11 PM | #1 |
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Electrochemistry cathode question
I am currently looking into photocatalytic materials. Now I might be over simplifying this process but it my observation that I am just looking at electrolysis but instead of electricity provided by a battery I am using electron hole pairs to generate a potential to do my splitting. My question is about the cathode side. If I am splitting distilled water what happens at the cathode to the OH radicals if they have no impurities in the water do combine back with the hydrogen and form water or do they just stay in solution turning it more basic. It just would be easier and cheaper to be able to determine the amount of O2 and H2 produced from being able to determine the amount of OH in solution
Cathode (reduction): 2 H2O(l) + 2e− → H2(g) + 2 OH-(aq) |
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| Nov20-12, 02:28 AM | #2 |
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So far you used electron - what about hole?
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| Nov26-12, 12:08 PM | #3 |
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The hole is taken care of by the Anode region
Anode (oxidation): 2 H2O(l) → O2(g) + 4 H+(aq) + 4e− So we have a net of 2 electrons the formation of H2 and O2 gas but what happens to the H+ ions and OH- ions do the recombine to from water, hydrogen peroxide, H3O. In normal electrolysis they have a bind with the soluble materials to make the water conduct but in my example there are no other ions present. So in a traditional electrolysis problem the reaction would not take place cause there would be the inability of distilled water to conduct but here where are talking about cathodes and anodes microns apart so there must be some sort of interaction over such a small distance. |
| Nov26-12, 12:51 PM | #4 |
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Electrochemistry cathode question
H+ + OH- -> H2O
High resistance doesn't mean electrolysis doesn't take place, it just slows it down and forces you to use much higher voltages to get some reasonable reaction rate. |
| Nov26-12, 01:05 PM | #5 |
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That's what I was afraid of. I was hoping that there would be an easier method into looking at the production of Hydrogen and Oxygen with out using a gas spectrometer. I was hoping for the OH radicals to remain and I could use a simple PH meter to indirectly get how much H2 was being produced
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| Nov26-12, 01:18 PM | #6 |
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OH radicals don't change solution pH. OH- anions do.
Looks like you are trying to do things you have no idea about. |
| Nov26-12, 01:36 PM | #7 |
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You are right I have neither training in electrochemistry nor photochemistry in fact I am a physics undergrad. I was trying to get some insight on the reactions that are taking place on the surface of my material I am working with. I think however I that what I meant by OH radicals is in fact OH anions which are formed by the oxidation and reduction reactions that take place at the surface. I just wanted a way to get a quantitative method of looking at hydrogen production without gas spectrometer since I can’t get ahold of one.
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| Nov26-12, 03:51 PM | #8 |
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Anything wrong with the volume measurement?
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| Nov26-12, 04:52 PM | #9 |
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The reaction produces μmol and it wont give me the Hydrogen production rates
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| Nov26-12, 05:06 PM | #10 |
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OK.
What about measuring the charge? That would require an assumption that electrode reaction is 100% efficient, but even if it is not, efficiency should be more or less constant all the time. |
| Nov26-12, 05:18 PM | #11 |
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How would I go about that and I assume you are talking about the 2 electron difference from cathode to anode
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| Nov26-12, 05:38 PM | #12 |
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No, amount of gas produced at each electrode is directly proportional to the charge that passed through the cell.
http://en.wikipedia.org/wiki/Faraday...f_electrolysis |
| Nov26-12, 06:31 PM | #13 |
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Would this even be possible to measure this on a Nano-scale? I know how many electronvolts I am putting into the system but so far current research says there is only about 7%-15% efficiency. Also different doping agents I add to the system lower or raise the band gap of the material and effect the production of Hydrogen. I think I am going to have to ask the research group to let me borrow one of their gas spectrometers. I was just hoping that this OH anions and PH would make it a lot easier to get relationships between the reactions.
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| Nov27-12, 02:00 AM | #14 |
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2H2O -> 2H2 + O2 Overall there are no OH- anions produced in the system and pH stays constant. |
| Nov29-12, 11:04 AM | #15 |
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Well what about a pressure measurement. Given that there should be no formation of other gas could I use a change in pressure to tell the hydrogen amount. Based upon partial pressure and we have a a 2:1 relationship between the hydrogen and oxygen. I think I would have to assume Ideal gas but it would give me some place to start so i can design a qualitative experiment and then work on the quantitative one.
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| Nov29-12, 11:29 AM | #16 |
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Pressure on the nano scale? No idea if it is doable (I know nothing about the subject). However, I suppose pressure buildup is not something you want, as when the pressure goes up it changes the potential at which the reaction goes. That will interfere with the kinetics that you want to measure.
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